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Určete Taylorův polynom druhého stupně funkce

$f(x)=\sin(3x+\frac{\pi}{4})+5.$ f(x)=\sin(3x+\frac{\pi}{4})+5. se středem v bodě $a=-\frac{\pi}{6}$ a=-\frac{\pi}{6}.

$5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})-\frac{9\sqrt 2}{4}(x+\frac{\pi}{6})^2$ 5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})-\frac{9\sqrt 2}{4}(x+\frac{\pi}{6})^2

$5+\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 2}{4}(x+\frac{\pi}{6})^2$ 5+\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 2}{4}(x+\frac{\pi}{6})^2

$5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 2}{4}(x+\frac{\pi}{6})^2$ 5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 2}{4}(x+\frac{\pi}{6})^2

$5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 2}{2}(x+\frac{\pi}{6})^2$ 5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 2}{2}(x+\frac{\pi}{6})^2

$5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 2}{2}(x-\frac{\pi}{6})^2$ 5-\frac{\sqrt 2}{2}+\frac{3\sqrt 2}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 2}{2}(x-\frac{\pi}{6})^2

$5-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{\sqrt 2}{4}(x+\frac{\pi}{6})^2$ 5-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}(x+\frac{\pi}{6})+\frac{\sqrt 2}{4}(x+\frac{\pi}{6})^2

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Matematika - LS 24/25

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